It can be anything, anything at all. The answer just needs to equal 7 (not 17, 27, 37, etc.)
-
Yes, because you can use DeMoivres Theorom to represent that as cos(pi)+ i(sin(pi)), sin(pi)=0, cos(pi)= -1, -1 +8=7. And also, .999...+6=7 Look up the full proof on wikipedia, cause im lazy. [Edited on 06.12.2011 8:26 PM PDT]
-
-93 + 100 = 7
-
Cube root of 343.
-
[quote][b]Posted by:[/b] Bearded Elf e^(pi*i) + 8[/quote] does that actually work since you would have an imaginary number don't feel like figuring it out
-
e^(pi*i) + 8
-
[quote][b]Posted by:[/b] IcedToastcat 7x1 [u]0x7[/u] 8-1 10-3 11-4 12-5 5+2 3+4 4+5-2[/quote] *cough*
-
(7x7x7x7x7x7x7x7x7x7x7x7x7x7x7x7x7x7x7x7x0)+7=7
-
343^1/3
-
7X7(radical)-7+7=7
-
7x1 0x7 8-1 10-3 11-4 12-5 5+2 3+4 4+5-2
-
I do not feel like doing math right now, but I will give you ONe and only ONE: 98-91=7
-
(afdfa7gca8chg6gagg736g8ca6hchiv6avggyisdauif)^0 -1+7 [Edited on 06.12.2011 8:11 PM PDT]
-
[quote][b]Posted by:[/b] the panzie man Infinite equations that represent 7 = 7[/quote]/thread
-
(radical 7)squared
-
Infinite equations that represent 7 = 7
-
77-70 67-60 57-50 123456787-123456780
-
3.5 x 2= 7 49/ 7= 7 9 - 2= 7 Easy.
-
1 x 7 3.5 x 2 2.333... x 3 1.75 x 4 1.4 x 5 1.1666... x 6
-
1 x 7= 7 [Edited on 06.12.2011 8:06 PM PDT]
-
1+6 2+5 3+4 2+3 8-1 9-2 10-3 ...
-
7+0 = 7 1 + 6 = 7 2 + 5 = 7 3 + 4 = 7 4 + 3 = 7 5 + 2 = 7 -7+14=7 this isn't hard
-
12x12=7